Let $\mathfrak{g}$ be a semisimple lie algebra. Let $\alpha_1,.., \alpha_n$ form a simple root system, and let the corresponding system of fundamental weights be $\lambda_1,..,\lambda_n$. I have been told that $\frac{1}{2} \sum_{\alpha \in \Delta^+} \alpha = \lambda_1 + … + \lambda_n$.
However, I have difficulty in seeing why this can be true. For example, in the sl(3) case, the positive roots are $e_1 – e_2, e_2 – e_3, e_1 – e_3$, where $e_i \in \mathfrak{h}^*$ is the functional which pulls out the $i$'th diagonal element. $\frac{1}{2} \sum_{\alpha \in \Delta^+} \alpha = e_1 – e_3$. However, in this case, if we take the simple root system $\alpha_1 = e_1 – e_2, \alpha_2 = e_2 – e_3$, then $\lambda_i = e_1 + .. + e_i$, and so $\lambda_1 + \lambda_2 = 2e_1 + e_2$. Where does my reasoning go wrong in the sl(3) case?
And why is it true that the sum of fundamental weights is equal to half the sum of positive roots?
Best Answer
Nothing is wrong with your reasoning for sl(3). Remember that in sl(3), there is the additional relation $e_1 + e_2 + e_3 = 0$, because the trace of a matrix in sl(3) is 0. So $e_1 - e_3 = 2e_1 + e_2$.